Prove $\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon}\int\limits_{X\leqslant \varepsilon}X\mathrm{d}\mathbb{P}=0$

Question

Let $X$ be a random variable taking values in $[0,+\infty]$. Prove that: $$\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon}\int\limits_{X\leqslant \varepsilon}X\mathrm{d}\mathbb{P}=0 ~~~~\mathrm{and~~~} \lim_{x \to +\infty}\frac{1}{x}\int\limits_{X\leqslant x}X\mathrm{d}\mathbb{P}=0.$$

Attempt. Of course we have $\displaystyle \frac{1}{\varepsilon}\int\limits_{X\leqslant \varepsilon}X\mathrm{d}\mathbb{P}\leqslant \frac{1}{\varepsilon} \varepsilon =1,$ but we don't get anything interesting by that (the same arguments holds for the other integral also).

Thank you in advance.

Answer 1

Observe that $$ \int_{X\leq \epsilon}X\ d\mathbb P=\int_{0<X\leq \epsilon}X\ d\mathbb P\leq \int_{0<X\leq \epsilon}\epsilon\ d\mathbb P=\epsilon \mathbb P(0<X\leq \epsilon), $$ and therefore $$ 0\leq \frac{1}{\epsilon}\int_{X\leq \epsilon}X\ d\mathbb P\leq \mathbb P(0<X\leq \epsilon). $$ As $\epsilon\to 0$ the right side tends to $\mathbb P(0<X\leq 0)$ which is zero (regardless of whether $\mathbb P(X=0)>0$), so we get the desired limit, by the squeeze theorem.

The second limit has a simple proof if $\mathbb EX<\infty$, in which case $$ \int_{X\leq x}X\ d\mathbb P\leq \int_{\Omega}X\ d\mathbb P=\mathbb EX<\infty, $$ and therefore $$ \frac{1}{x}\int_{X\leq x}X\ d\mathbb P\leq\frac{\mathbb EX}{x} $$ which tends to zero as $x\to\infty$.

Here is a more involved argument which works when $\mathbb EX=\infty$ as well. Apply the tail formula for the expectation to the random variable $X\cdot 1_{X\leq x}$ to obtain the formula $$\int_{X\leq x}X\ d\mathbb P=\int_0^x\mathbb P(t\leq X<\infty)\ dt.$$ As $t\to\infty$ the quantity $\mathbb P(t\leq X<\infty)$ tends to zero (regardless of whether $\mathbb P(X=\infty)$ is positive). Thus, for every $\epsilon>0$ we can find an $N<\infty$ such that $\mathbb P(t\leq X<\infty)<\epsilon$ for all $t>N$. Consequently for all $x>N$ we have the upper bound $$ \int_0^x\mathbb P(t\leq X<\infty)\ dt\leq N+\int_{N}^x\mathbb P(t\leq X<\infty)\ dt\leq N+(x-N)\epsilon. $$ Dividing both sides by $x$ and sending $x\to\infty$ yields that $$\limsup_{x\to\infty}\frac{1}{x}\int_{X\leq x}X\ d\mathbb P\leq\epsilon.$$ Since $\epsilon>0$ was arbitrary, we therefore obtain that $$ \limsup_{x\to\infty}\frac{1}{x}\int_{X\leq x}X\ d\mathbb P=0, $$ and hence the desired limit.